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olganol [36]
3 years ago
8

Translate the following statement into an algebraic expression: Use x for your variable. The sum of the square of a number and e

leven times the number.​

Mathematics
2 answers:
dusya [7]3 years ago
6 0

Step-by-step explanation:

Let x be the number.

Then the algebraic expression will be x^2 + 11x.

Setler79 [48]3 years ago
6 0
Hope this will help you

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Please Explain!!<br> I need help plss
Nimfa-mama [501]

Answer:

See step by step

Step-by-step explanation:

For A. use any two factors that multiply to 100 that isn't 25 and 4.

For b Use one obtuse angle (angle over 90 degree) and two acute angles. Make sure they add up to 180.

For c Use any two values that add up to 52 since they didnt establish y as the midpoint.

For d. measure of angle 1 is 45 and measure of angle 2 is 45 degrees

6 0
3 years ago
Read 2 more answers
1. Jemima wants to make chocolate chip walnut brownles. Chocolate chips come in a 12oz bag that costs $3. Walnuts
fomenos

Answer:

4 bags of chocolate chips and 1 bag of walnuts.

Step-by-step explanation:

Find how many ounces are in 3 pounds:

16 oz = 1 lb

16 oz × 3 = 3 lbs

48 oz = 3 lbs

Variable x = number of chocolate chips

Variable y = number of walnuts

4x + 1y = 15

Keep in mind it must equal up to 48 oz and be $15 or under.

4 bags × 12 oz = 48 oz

1 bag × 4 oz = 4 oz

48 oz + 4 oz = 52 oz > 48 oz

4 bags × $3 = $12

1 bag × $2 = $2

$12 + $2 = $14 < $15

Both requirements have been met.

7 0
3 years ago
If a pair of standard dice are rolled, what is the probability that one of the dice will be a 3 and the other a 4?
eimsori [14]

Answer:

0.055

Step-by-step explanation:

file is attached with explanation

7 0
3 years ago
What belongs in the box marked with the question mark in the proof?
koban [17]
<span>the blank boxes are for you to plug in x=20 to prove its right. so it would be 3(20)-4=2(20+8) 60-4=2(28) 56=56 so its true!</span>
6 0
4 years ago
. Use the quadratic formula to solve each quadratic real equation. Round
Liono4ka [1.6K]

Answer:

A. No real solution

B. 5 and -1.5

C. 5.5

Step-by-step explanation:

The quadratic formula is:

\begin{array}{*{20}c} {\frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}}} \end{array}, with a being the x² term, b being the x term, and c being the constant.

Let's solve for a.

\begin{array}{*{20}c} {\frac{{ 5 \pm \sqrt {5^2 - 4\cdot1\cdot11} }}{{2\cdot1}}} \end{array}

\begin{array}{*{20}c} {\frac{{ 5 \pm \sqrt {25 - 44} }}{{2}}} \end{array}

\begin{array}{*{20}c} {\frac{{ 5 \pm \sqrt {-19} }}{{2}}} \end{array}

We can't take the square root of a negative number, so A has no real solution.

Let's do B now.

\begin{array}{*{20}c} {\frac{{ 7 \pm \sqrt {7^2 - 4\cdot-2\cdot15} }}{{2\cdot-2}}} \end{array}

\begin{array}{*{20}c} {\frac{{ 7 \pm \sqrt {49 + 120} }}{{-4}}} \end{array}

\begin{array}{*{20}c} {\frac{{ 7 \pm \sqrt {169} }}{{-4}}} \end{array}

\begin{array}{*{20}c} {\frac{{ 7 \pm 13 }}{{-4}}} \end{array}

\frac{7+13}{4} = 5\\\frac{7-13}{4}=-1.5

So B has two solutions of 5 and -1.5.

Now to C!

\begin{array}{*{20}c} {\frac{{ -(-44) \pm \sqrt {-44^2 - 4\cdot4\cdot121} }}{{2\cdot4}}} \end{array}

\begin{array}{*{20}c} {\frac{{ 44 \pm \sqrt {1936 - 1936} }}{{8}}} \end{array}

\begin{array}{*{20}c} {\frac{{ 44 \pm 0}}{{8}}} \end{array}

\frac{44}{8} = 5.5

So c has one solution: 5.5

Hope this helped (and I'm sorry I'm late!)

4 0
3 years ago
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